Consider the unbalanced chemical equation below:

CaSiO_{3}(*s*) + HF(*g*) → CaF_{2}(*aq*) + SiF_{4}(*g*) + H_{2}O(*l*)

Suppose a 32.9-g sample of CaSiO_{3} is reacted with 31.8 L of HF at 27.0°C and 1.00 atm. Assuming the reaction goes to completion, calculate the mass of the SiF_{4} and H_{2}O produced in the reaction.

We first need to balance the given chemical equation. We add a coefficient of 6 to HF to balance the number of F on both sides of the equation. We also add a coefficient of 3 to H_{2}O to balance the number of H and O. The balanced chemical equation is:

We then need to determine the moles of CaSiO_{3} and HF given. The molar mass of CaSiO_{3} is 40.08 g/mol Ca + 28.09 g/mol Si + 3(16.00 g/mol O) = **116.17 g/mol**. The moles of CaSiO_{3} given is:

We can calculate the moles of HF using the **ideal gas law**. We're given **P = ****1.00 atm**, **V = ****31.8 L**, and **T = 27.0°C = ****300.15 K**.

Since we're given the amounts of both reactants, we need to determine which between CaSiO_{3} and HF is the limiting reactant. The **limiting reactant** forms the least amount of products amount reactants and actually determines the amount of products formed.

The molar mass of SiF_{4} is 28.09 g/mol Si + 4(19.00 g/mol F) = **104.09 g/mol** and the molar mass of H_{2}O is 2(1.01 g/mol H) + 16.00 g/mol O = **18.02 g/mol**.

Gas Stoichiometry